Convex Optimization - affine Set

A set $A$ is said to be an affine set if for any two distinct points, the pne passing through these points pe in the set $A$.

Note −

$S$ is an affine set if and only if it contains every affine combination of its points.

Empty and singleton sets are both affine and convex set.

For example, solution of a pnear equation is an affine set.

Proof

Let S be the solution of a pnear equation.

By definition, $S=left { x in mathbb{R}^n:Ax=b ight }$

Let $x_1,x_2 in SRightarrow Ax_1=b$ and $Ax_2=b$

To prove : $Aleft [ heta x_1+left ( 1- heta ight )x_2 ight ]=b, forall heta inleft ( 0,1 ight )$

$Aleft [ heta x_1+left ( 1- heta ight )x_2 ight ]= heta Ax_1+left ( 1- heta ight )Ax_2= heta b+left ( 1- heta ight )b=b$

Thus S is an affine set.

Theorem

If $C$ is an affine set and $x_0 in C$, then the set $V= C-x_0=left { x-x_0:x in C ight }$ is a subspace of C.

Proof

Let $x_1,x_2 in V$

To show: $alpha x_1+eta x_2 in V$ for some $alpha,eta$

Now, $x_1+x_0 in C$ and $x_2+x_0 in C$ by definition of V

Now, $alpha x_1+eta x_2+x_0=alpha left ( x_1+x_0 ight )+eta left ( x_2+x_0 ight )+left ( 1-alpha -eta ight )x_0$

But $alpha left ( x_1+x_0 ight )+eta left ( x_2+x_0 ight )+left ( 1-alpha -eta ight )x_0 in C$ because C is an affine set.

Therefore, $alpha x_1+eta x_2 in V$

Hence proved.